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For dilute nonrelativistic nucleons in thermal equilibrium, their Maxwell-Boltzmann distribution gives $n_j=g_j(m_{j}kT/2\pi {\hslash }^2{)}^{3/2}\exp [({\mu }_j-m_j{c}^2)/kT]$. The neutron and proton spin degeneracies are equal, and their mass prefactor ratio is nearly one. ThusWeak reaction equilibrium imposes ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. Charge neutrality in an electron-positron plasma, together with the small baryon asymmetry, makes ${\mu }_e/kT$ very small. Assuming negligible neutrino lepton asymmetry also makes ${\mu }_{{\nu }_e}/kT$ small. Hence the difference, rather than the absolute baryon chemical potentials, is negligible, yieldingA large neutrino chemical potential would invalidate this last approximation; it is an assumption about the lepton asymmetry, not a general rule that baryonic chemical potentials vanish.

The weak-conversion rate per expansion time falls as $T^3$. Below the stated threshold it becomes less than one, so neutron-proton interconversion cannot track the rapidly changing equilibrium ratio: cosmological weak freeze-out occurs. Free neutrons subsequently decay, reducing their abundance until nuclei can survive photodissociation. Once the deuterium bottleneck is passed, nuclear reactions rapidly assemble neutrons into the strongly bound helium-4 nucleus. The short available time, low density and missing stable mass-five and mass-eight nuclei impede extensive production of heavier nuclei, leaving nearly all surviving neutrons in helium.

If $r=n_n/n_p$ just before helium formation, each helium nucleus uses two neutrons and two protons, so $n_{\mathrm{He}}=n_n/2$. Neglecting binding-energy corrections to nuclear masses,The remaining protons mainly become hydrogen. Counting all four helium nucleons, rather than only the captured neutrons, is essential to this mass fraction.

Chemical equilibrium for $p+e^{-}\leftrightarrow H+\gamma$ implies ${\mu }_H={\mu }_p+{\mu }_e$, because equilibrium photons have zero chemical potential. Dividing the three nonrelativistic Maxwell-Boltzmann number densities therefore givesThe Hydrogen binding energy is $I=m_p+m_e-m_H=13.6\mathrm{eV}$ in natural units. For ground-state hydrogen, with hyperfine splittings unresolved, $g_H=4$ and $g_p=g_e=2$, so the degeneracy ratio is one. Equivalently one can omit nuclear-spin degeneracy consistently from both the proton and atom. Excited hydrogen levels are exponentially suppressed at recombination temperatures. Also $m_H/m_p=1+O(m_e/m_p)$, so its effect in the prefactor is negligible. ThusThe sign of the exponential follows from the smaller rest mass of the bound atom: cooling favours neutral hydrogen once translational entropy no longer compensates the binding energy.

Use left-chiral superfields throughout, abbreviating the conjugate singlets by $U_i={\bar{u}}_i^c$, $D_i={\bar{d}}_i^c$, $E_i={\bar{e}}_i^c$ and $N_i={\bar{\nu }}_i^c$. The printed hypercharge convention is the negative of the usual one, so $Q_{\mathrm{em}}=T_3-Y$; consequently $H_1$ is the down-type and $H_2$ the up-type Higgs doublet. Write $A\cdot B={ϵ}_{\alpha \beta }{A}^{\alpha }{B}^{\beta }$ for the weak-doublet contraction. Assign baryon number $B(Q)=1/3$, $B(U)=B(D)=-1/3$, and lepton number $L(L)=1$, $L(E)=L(N)=-1$; both Higgs fields have $B=L=0$.

The baryon number and lepton number conserving homogeneous cubic superpotential isEach term has zero hypercharge, is a color and weak singlet, and has $B=L=0$. Higgs expectation values give the usual quark and charged-lepton Yukawa interaction masses and, here, Dirac neutrino masses. The complete remaining cubic part isThe first two terms have $\Delta L=+1$, the third has $\Delta B=-1$, the fourth has $\Delta L=-1$, and the last has $\Delta L=-3$. Their Hermitian conjugates reverse these signs. Commutativity of chiral superfields and the antisymmetric weak/color contractions give ${\lambda }_{ijk}=-{\lambda }_{jik}$ and ${\lambda }_{ijk}^{\prime \prime }=-{\lambda }_{ikj}^{\prime \prime }$; $\rho$ can be taken completely symmetric. All other coupling tensors are unrestricted complex tensors.

To check completeness, a cubic color singlet containing a quark doublet and an antiquark singlet must contain a colorless weak doublet; the allowed hypercharges give $Q{H}_{2}U$, $Q{H}_{1}D$ and $QLD$. Three color-antitriplet singlets can form a singlet with ${ϵ}^{abc}$, but hypercharge permits only $UDD$. Three quark doublets cannot be a weak singlet. With no colored fields, an even number of weak doublets is necessary. Two doublets and a singlet give precisely $L{H}_{1}E$, $LLE$, $L{H}_{2}N$ and $H_1{H}_{2}N$; contractions of two identical Higgs superfields vanish. With no doublets the only neutral triple is $NNN$. Thus $W_3=W_{3,\mathrm{cons}}+W_{3,\mathrm{viol}}$ is the most general homogeneous cubic superpotential for the printed fields, including their singlet neutrinos.

If “cubic” is intended to mean a polynomial of degree at most three, also addThe $\mu$ term conserves both charges; the other nonconstant terms have respectively $\Delta L=1,-2,-1$. This distinction will matter for the printed claim about R-parity.

For squark-mediated proton decay, choose ${\lambda }_{11k}^{\prime }$ and ${\lambda }_{11k}^{\prime \prime }$ with $k=2$ or $3$. The same down-type singlet squark can appear in both vertices; $k=1$ would give ${\lambda }_{111}^{\prime \prime }=0$ by antisymmetry. The first term includes an electron/up-quark/squark vertex, and the second an up-quark/down-quark/squark vertex. Exchanging a squark of mass $M$ and expanding its propagator at momenta much smaller than $M$ produces the dimension-six effective operatorThe left- and right-handed pairs are separately contracted spinor bilinears. This operator converts an up/down quark pair into a positron and an antiquark; the spectator up quark can combine with the antiquark into the neutral pion. Equivalently its hadronic matrix element between a proton and a pion is permitted, so it induces $p\to e^{+}{\pi }^0$. Having both the baryon number and lepton number violating vertices is what enables this channel.

Since $C$ has mass dimension $-2$, dimensional analysis givesThe dimensionless $c_h$ includes the hadronic matrix element and two-body phase space; the pion mass changes this factor rather than the leading dimensions. Dimensional grounds alone do not give a precise hadronic coefficient. Using the lifetime supplied in the paper, rather than substituting a different bound, gives $\Gamma <1/(2.4\times 10^{64})\mathrm{GeV}=4.17\times 10^{-65}\mathrm{GeV}$ and henceFor an illustrative $100\mathrm{GeV}$ squark this is about $6.5\times 10^{-29}{c}_h^{-1/2}$ when $m_p$ is rounded to $1\mathrm{GeV}$. The bound constrains the relevant flavour product and depends on the exchanged mass; separate smallness of each coupling does not follow without an additional assumption.

Define matter parity and R-parity byEvery matter superfield $Q,U,D,L,E,N$ has $P_M=-1$, whereas $H_1,H_2$ and the vector superfields have $P_M=+1$. For ordinary quarks and antiquarks, $3(B-L)=\pm 1$ is odd and $2s=1$, so $R_p=+1$. For charged leptons, neutrinos and their antiparticles, $3(B-L)=\mp 3$ is odd and $2s=1$, again giving $R_p=+1$; the singlet neutrino fermion obeys the same rule. Gluons, the photon, $W$ and $Z$ have $B=L=0$ and spin one, hence $R_p=+1$; Higgs bosons have $B=L=s=0$ and also $R_p=+1$. Thus every Standard Model particle is R-even, including the corresponding antiparticles. Changing spin by one half within a supermultiplet reverses $R_p$, so squarks, sleptons, gauginos and Higgsinos are R-odd.

Each conserving cubic term contains two R-odd matter superfields and one even Higgs superfield, so it is even under matter parity. Each term in $W_{3,\mathrm{viol}}$ has an odd number of matter superfields: three for $LLE,LQD,UDD,NNN$, and one for $N{H}_1{H}_2$. All are therefore forbidden. At the component level the two fermion spin factors in a Yukawa interaction multiply to $+1$, so the same selection rule follows directly from R-parity. This verifies all cubic baryon/lepton-number violating terms are forbidden while the Yukawa fermion mass terms survive.

The blanket assertion needs a qualification beyond homogeneous cubic terms. The Higgs $\mu$ term is even; $L{H}_2$ and $N$ are odd and forbidden. But $N_i{N}_j$ is even and permits a Majorana neutrino mass despite violating lepton number by two. Likewise the higher-dimensional operators $QQQL$ and $(L{H}_2{)}^2$ are even. Consequently R-parity alone does not enforce all-order conservation of baryon number and lepton number, or absolute proton stability; the stated precise exclusion holds for the violating cubic interactions above.

Two physical implications follow from multiplicative conservation of R-parity. Superpartners are produced in even numbers from an initial state consisting only of Standard Model particles, leading to pair production at the lowest threshold and parity-conserving cascades. The lightest supersymmetric particle is stable, since one R-odd particle cannot decay exclusively into R-even particles and no lighter R-odd state is available. A neutral weakly interacting lightest supersymmetric particle can therefore be a dark-matter candidate and produce missing-energy signatures; neutrality and a suitable abundance require further model assumptions.

For $p+e\rightleftharpoons H+\gamma$, chemical equilibrium and the vanishing photon chemical potential imply ${\mu }_H={\mu }_p+{\mu }_e$. Taking the ratio of the three nonrelativistic Maxwell--Boltzmann number densities cancels these chemical potentials:The ground-state hydrogen atom has four combined electron and proton spin states if the small hyperfine splitting is ignored, whereas $g_p=g_e=2$. Hence $g_H/(g_p{g}_e)=1$. With Hydrogen binding energy $B=M_p+M_e-M_H$ and $M_H\simeq M_p$, the Saha ionization equation givesThe approximation in the prefactor neglects relative corrections of order $(M_e+B)/M_p$; the Hydrogen binding energy must be retained in the exponential.

The nucleons are nonrelativistic, so the Maxwell-Boltzmann distribution givesBoth spin degeneracies are two. Taking the ratio and using the specified negligible chemical potentials yields the neutron-to-proton ratioThe mass prefactor differs from unity only at the per-mille level. The dependence on $T_F$ is much stronger: $d\ln r_F/d\ln T_F=Q/T_F>0$. Thus earlier, hotter weak freeze-out leaves more neutrons relative to protons. For illustration, $T_F=1\mathrm{MeV}$ gives $r_F\simeq 0.28$; $T_F=0.8\mathrm{MeV}$ gives $r_F\simeq 0.20$. Free-neutron decay after freeze-out lowers the ratio further before most Big Bang nucleosynthesis takes place.

The small baryon-to-photon ratio suppresses nuclear equilibrium abundances through the factor ${\eta }^{A-1}$. Cooling eventually makes the binding-energy exponential favourable, but the deuterium bottleneck delays substantial nuclear burning until well after cosmological weak freeze-out. In ordinary Big Bang nucleosynthesis, nearly every neutron surviving that delay ends up in helium-4. This makes the final primordial helium mass fraction chiefly a neutron budget rather than an unrestricted equilibrium abundance.

Let $r_F$ be the neutron-to-proton ratio at freeze-out, let $\Delta t$ be the delay until effective nuclear burning, and let ${\tau }_n$ be the neutron mean lifetime. The initial fraction of baryons which are neutrons is $x_{n,F}=r_F/(1+r_F)$. Because neutron decay converts a neutron into a proton, total baryon number remains constant, while the neutron fraction becomes $x_{n,B}=x_{n,F}{e}^{-\Delta t/{\tau }_n}$. Each helium nucleus uses two neutrons and four baryons. Hence neutron-limited helium synthesis givesEquivalently, $Y_4=2r_B/(1+r_B)$ with $r_B=x_{n,B}/(1-x_{n,B})$. This accounts for the extra protons produced by neutron decay; multiplying $r_F$ alone by the survival factor would miss their contribution. The following responses refer to ordinary neutron-limited burning, without exotic particle injection or a change to the nuclear network.

Let $n_n$ and $n_p$ count the available neutrons and protons immediately before their incorporation into nuclei. In the usual neutron-limited helium synthesis approximation, all surviving neutrons go into ${}^4\mathrm{He}$ and the remaining protons form ${}^1\mathrm{H}$. Each helium nucleus contains two neutrons and two protons, soUsing the equal-nucleon-mass approximation and neglecting trace nuclei and binding-energy corrections, the primordial helium mass fraction isThe ratio is the pre-assembly nucleon ratio, not the number ratio of free hydrogen nuclei to helium nuclei afterward. The derivation assumes enough protons to bind all the neutrons, as in standard Big Bang nucleosynthesis. The familiar $n_p/n_n\simeq 7$ gives $Y_p\simeq 1/4$ and hydrogen mass fraction $X_p\simeq 3/4$.

Write $n_{56}$ and $n_{54}$ for the number densities of nickel-56 and iron-54. Both nuclei contain 28 neutrons, while their proton numbers differ by two. Dividing the positive-exponential nuclear statistical equilibrium formulas printed in the PDF therefore gives the nickel–iron nuclear equilibrium ratioThe neutron-density factors cancel. $K$ contains the isotope-dependent statistical and translational mass factors suppressed in the supplied proportionality; it is constant in this simplified calculation. The TeX conversion incorrectly reverses the exponential sign, so the PDF sign is essential here.

In the specified restricted composition, equality of total proton and neutron numbers meansConsequentlyDifferentiation, with the suppressed prefactors held constant, givesUsing the paper's supplied signed energy difference producesWith that negative value, $F$ tends to zero at both temperature endpoints, increases below this temperature and decreases above it. At the stationary point $d^2\log F/d{T}^2=-3/T^2<0$, so it is indeed a maximum for the printed toy data.

There is a genuine physical-data qualification: the stated negative energy difference is inconsistent with interpreting $Q$ as the conventional positive total nuclear binding energy of these actual isotopes. Using atomic masses, whose electron masses cancel in this reaction,The masses used are $1.00782503223\mathrm{u}$, $53.93960899\mathrm{u}$ and $55.94212855\mathrm{u}$, from NIST hydrogen masses, NIST iron masses and NIST nickel masses. Small electronic binding corrections do not change the sign. With a positive $\Delta B$, the simplified $F(T)$ above is strictly decreasing for $T>0$ and has no finite maximum. Thus the numerical maximum follows from the supplied negative parameter, not from consistent physical binding energies; changing the TeX sign to recover a maximum would not repair the PDF's data.

Nuclear statistical equilibrium is appropriate when the relevant forward and inverse strong or electromagnetic reactions are fast compared with changes of temperature, density and composition. It is useful in sufficiently hot late silicon-burning material and in the hottest explosive burning regions of massive stars. Some burning zones reach only quasi-equilibrium, and cooling or expansion eventually freezes the abundances out; an equilibrium expression cannot by itself predict the full ashes when these timescale conditions fail. Temperature, density, conserved baryon number and the electron fraction in stellar matter must all be specified. Negligible free-particle contributions in bulk bookkeeping are not the same as exactly zero free densities in the equilibrium formula.

Strong and electromagnetic reactions preserve total charge and baryon number, so they conserve the proton-to-neutron ratio. Weak reactions need not: electron capture and beta-plus decay convert protons into neutrons and lower it. Such conversions occur in hydrogen-burning networks and in the nitrogen-to-neon chain during helium burning. At high density in advanced stellar cores and core collapse, electron capture becomes particularly important; neutrino emission also affects energy and lepton loss. Beta-minus decay changes the ratio in the opposite direction and can compete during later relaxation. Nuclear equilibrium for strong reactions does not automatically imply weak equilibrium or a fixed proton-to-neutron ratio throughout all stellar evolution.

Use $f=X_{22}$ for the neon-22 mass fraction, temporarily keeping it distinct from the earlier metallicity. Let $n_b=\rho /m_u$ be the baryon density, ignoring small nuclear mass defects. Equal masses of carbon and oxygen giveCarbon-12 and oxygen-16 each have equal proton and neutron numbers, while neon-22 has ten protons and twelve neutrons. The total proton and neutron numbers per baryon are consequentlyHenceThe equal carbon/oxygen mass split cancels from this result: any mixture of these two symmetric nuclei has the same electron fraction. Taking the paper's metallicity symbol to mean $f$ gives its requested expression.

The literal pre-helium-burning nitrogen interpretation, however, needs a correction. Each initial nitrogen-14 nucleus receives two alpha particles, including eight additional baryons, before becoming neon-22; the intervening beta-plus decay also emits a neutrino. Thus, if the entire original metallicity is nitrogen-14 with mass fraction $Z_{\mathrm{initial}}$, conservation of the number of seed nuclei givesThis is neon-22 neutron excess from nitrogen-14. The two expressions are compatible only when their arguments are distinguished: the displayed $(11-Z)/(11+Z)$ law uses the final neon mass fraction, not literally the earlier nitrogen mass fraction. Treating those fractions as equal is an extra simplifying approximation, not a consequence of the stated capture chain.

A radioactively powered Type Ia supernova light curve rises to a broad maximum over roughly weeks as energy deposited in expanding ejecta diffuses out, then declines over the following weeks and months. Its power comes predominantly from the nickel-56 decay chain, first nickel to cobalt and then cobalt to iron. Gamma rays and charged decay products deposit heat, which is reprocessed into optical and infrared radiation. At later times, declining cobalt decay power and increasing gamma-ray escape shape the tail. It is not sustained nuclear fusion in the expanding ejecta. The amount of radioactive nickel, photon diffusion time and deposition efficiency all matter, so an uncorrected fixed luminosity is only an idealization.

For the requested two-isotope yield calculation, assume the same total processed iron-group mass, namely $1M_{\odot }$ as inferred from the zero-neon reference. Treat the final freeze-out ashes as nickel-56 and iron-54, neglecting residual light particles and additional weak neutronization. If $w$ is the nickel mass fraction, conservation of electron fraction givesSince $26/54=1/2-1/54$, this reduces to the metallicity dependence of nickel-56 yieldUnder the paper's intended identification $f=Z=1/50$,If instead $Z=1/50$ denotes the earlier nitrogen mass fraction, retaining the actual capture stoichiometry gives $w=1-27Z/7=323/350$ and $M_{56}\simeq 0.92M_{\odot }$. The same temperature and density do not alone fix a total burned mass; equality of that mass is also implicit in the comparison. The two-isotope assumption is a bulk-ash approximation, not a complete equilibrium network with all free particles identically absent.

For otherwise comparable explosions, approximate the peak luminosity by a quantity proportional to the radioactive nickel mass. Let $d$ be the true distance to an extremely low-neon event, and use the solar calibration $L_{\odot }/L_0=523/550$. Its measured flux is $F=L_0/(4\pi d^2)$, but the inferred standard-candle distance isThe source is intrinsically brighter than the calibration, so the distance is underestimated by about $2.5\%$, conventionally described here as about $3\%$. Equivalently, $d/d_{\mathrm{est}}=\sqrt{550/523}\simeq 1.0255$, consistent with the supplied rounded factor $1.03$. Using the literal nitrogen metallicity instead gives about a $4\%$ underestimate. These are conditional composition effects in this simplified model; real Type Ia distance estimation also standardizes light-curve shape and color.

Use an incoming spacelike momentum transfer $q$, so the physical absorption process has $p_X=P+q$, $Q^2=-q^2>0$, and $P\cdot q>0$. There is a sign inconsistency in the original PDF: its delta function has $P-p_X-q$, while its positive Bjorken x definition and target formulas use incoming $q$. With that literal delta function and $x>0$, the final-state invariant would bewhich becomes negative in the stated high-energy limit and cannot represent a physical hadronic final state. We therefore use ${\delta }^4(P+q-p_X)$, preserving the printed $x=Q^2/(2P\cdot q)$ and structure-function conventions. Equivalently, keeping an outgoing momentum variable requires reversing the signs in those definitions as well.

Assume an unpolarized target, incoherent scattering from effectively free spin-one-half partons, and the massless collinear parton approximation. In the Bjorken scaling limit, take $Q^2,P\cdot q\to \infty$ at fixed $0<x<1$, neglecting target and quark masses and intrinsic transverse momentum relative to $Q$. Write the incoming struck-quark momentum as $k=\xi P$ in the leading lightlike-reference approximation. Let $f_f(\xi )$ and ${\bar{f}}_f(\xi )$ be quark and antiquark parton distribution functions, counting all colours. For the naive scaling result also neglect radiative scaling violations; in full QCD these distributions depend logarithmically on the hard scale.

The outgoing struck quark has momentum $k^{\prime }=k+q$. Its on-shell condition givesAverage its initial spin and sum its final spin. The spin-averaged electromagnetic parton tensor has numeratorThe supplied three-gamma identity gives this directly: contract its middle index with $(k+q{)}_{\lambda }$ and trace with $/k$; the ${\gamma }_5$ trace vanishes. Thus no antisymmetric term survives the unpolarized spin average.

Define transverse tensorsOn the cut, $2k\cdot q=Q^2$, so $\tilde{k}=k+q/2=\xi \tilde{P}$. Expanding the two transverse terms givesThis also explicitly verifies electromagnetic current conservation, $q_{\mu }{S}^{\mu \nu }=q_{\nu }{S}^{\mu \nu }=0$.

The one-particle final-state Lorentz-invariant phase-space measure obeysIncluding the current's charge factor and the given $1/(4\pi )$ normalization, a single quark therefore contributesAn antiquark has the same unpolarized tensor and the same squared charge $Q_f^2$.

The parton convolution normalization for a hadronic tensor isThe $1/\xi$ is important: the distributions count parton numbers, whereas covariantly normalized one-particle responses carry the state normalization $2k^0$. The incoherent sum of normalized responses is $W/(2P^0)={\sum }_f\int d\xi (f_f+{\bar{f}}_f)w_f/(2k^0)$; in the collinear frame $k^0=\xi P^0$, giving the factor above. There is no extra colour multiplicity because it is already included in the distributions.

On the physical cut,Comparing the two transverse coefficients in the hadronic tensor gives, with $n_f=f_f+{\bar{f}}_f$,Therefore the requested deep-inelastic structure functions areThe notation $f_f$ here is the question's $f$ with its flavour index made explicit. The calculation proves the Callan-Gross relation, $F_2=2x{F}_1$, rather than merely citing it. Finite masses and radiative corrections modify this leading approximation; at leading QCD order the same formulas use scale-dependent parton distribution functions.

The quark number sum rule follows because a distribution is a number density. Integrating $f_f$ counts quarks and integrating ${\bar{f}}_f$ counts antiquarks; their difference is the net conserved flavour charge:Every sea quark-antiquark pair contributes zero to this difference. More formally the flavour-vector charge counts quarks minus antiquarks and has a fixed eigenvalue on the hadron state. Thus $N_f$ is the valence or net quark number, not the total number including the sea.

Use both final assumptions printed in the question: only up/down flavours and no antiquarks. The second does not follow just from restricting the flavour list. Then $\int f_u=N_u$, $\int f_d=N_d$, with squared charges $Q_u^2=4/9$ and $Q_d^2=1/9$. The proton has $(N_u,N_d)=(2,1)$ and the neutron $(1,2)$. Hence the valence-only proton and neutron structure-function moments areThe integral limits assume the scaling approximation is valid for the indicated weighted moments, for example by dominated convergence of the weighted integrands. The same net-quark sum rule alone would not fix these individual moments in the presence of antiquarks, since $F_2$ involves the sum of quark and antiquark distributions.

For a nonrelativistic particle, $E(p)=m{c}^2+p^2/(2m)$. The Gaussian integral yieldsIndeed ${\int }_0^{\infty }{p}^2{e}^{-p^2/(2m{k}_{B}T)}dp=\sqrt{\pi }(2m{k}_{B}T{)}^{3/2}/4$, which combines with $4\pi /h^3$ to give the prefactor. For the two species this givesChemical equilibrium imposes ${\mu }_a+{\mu }_{\alpha }={\mu }_b+{\mu }_{\beta }$. If the two massless bath species have zero chemical potentials, then ${\mu }_a={\mu }_b$; their masses alone do not imply this chemical-potential condition.

Weak reactions such as $n+{\nu }_e\leftrightarrow p+e^{-}$, with leptons treated as relativistic, interconvert neutrons and protons. Negligible lepton chemical potentials give approximately $n_n/n_p=\exp [-(m_n-m_p)c^2/(k_{B}T)]$, since their degeneracies and masses are nearly equal. Equilibrium favors protons increasingly as temperature falls. But the weak reaction rate decreases faster than the expansion rate and eventually falls below it: cosmological weak freeze-out prevents the ratio from following its equilibrium exponential to zero. Free neutrons subsequently decay, reducing the frozen ratio further.

In Big Bang nucleosynthesis, cooling eventually makes deuterium survive photodissociation. Nuclear reactions then build helium through deuterium and the light intermediate nuclei, locking almost all surviving neutrons into stable helium-four. Neglecting binding-energy corrections and taking $r=n_n/n_p\le 1$, each helium nucleus uses two neutrons and two protons, so $n_{\mathrm{He}}=n_n/2$ andFor example $r=1/7$ gives a helium mass fraction $1/4$, with the remaining baryons predominantly hydrogen.

Photon decoupling is a dynamical loss of frequent scattering, distinct from the chemical conversion of charged particles into neutral atoms. Before recombination, Thomson scattering on free electrons keeps photons coupled to the baryon plasma. The physical scattering rate isin units with $c=1$. Define the approximate decoupling temperature byEquivalently the photon mean scattering time becomes comparable to an expansion time. A refined last-scattering definition uses the optical depth and the maximum of the visibility function; it need not coincide exactly with the local rate criterion. As $X_e$ falls, scattering becomes inefficient, ordinarily after the half-ionized recombination stage, so $T_{\mathrm{dec}}<T_{\mathrm{rec}}$.

Residual electron freeze-out concerns the remaining ionized fraction. The Saha equation presumes sufficiently rapid forward and reverse reactions to maintain chemical equilibrium. Real recombination is slowed by radiative bottlenecks and expansion; the free-electron fraction can therefore exceed its equilibrium value. Eventually the effective recombination rate per free electron, roughly ${\alpha }_{\mathrm{eff}}(T)n_e$, falls below $H$. The remaining electrons cannot all find and recombine with protons within an expansion time. A small nonzero residual fraction survives while the Saha prediction decreases exponentially toward zero.

The sketch shows this lag and residual floor. Its kinetic curve and the position of the decoupling marker are schematic: the data supplied determine the equilibrium curve, but not a quantitative recombination history or an exact $T_{\mathrm{dec}}$. Determining those needs the expansion history and atomic transition rates. The horizontal axis decreases toward the right to follow cosmological cooling.

The supplied rate law gives a local logarithmic temperature sensitivity, evaluated at fixed number densities. Since $\eta \propto T^{-1/3}$,Thus a power law here describes the tangent logarithmic slope near the chosen temperature, not an exact power law over all temperatures. For the three reactions, the factors $A{Z}_i^2{Z}_j^2$ are $1/2$, $24$ and $192/7$, respectively. At $T_6=15$ the local temperature exponent of a thermonuclear reaction givesHence $\alpha \simeq 4$, $\beta \simeq 16$, $\gamma \simeq 16$ and $\gamma >\beta$. The larger reduced mass for reaction 34 produces the last inequality. Composition changes contribute separately to an evolving reaction rate; they are held fixed for this derivative.

For the effective proton-proton reaction network, let the event rates beThe factors of one half count identical pairs once. Event 11 consumes two protons and makes one deuteron; event 21 consumes a deuteron and a proton and makes helium-3; event 33 consumes two helium-3 nuclei and makes one helium-4 nucleus and two protons. Under the stipulated fast-capture approximation, event 34 consumes helium-3 and helium-4 and supplies one mass-seven nucleus, and event 17 consumes that lithium-7 nucleus and a proton and makes two helium-4 nuclei. ThereforeFor closure, ${\dot{n}}_7={\lambda }_{34}{n}_3{n}_4-{\lambda }_{17}{n}_1{n}_7$. The stoichiometry conserves $n_1+2n_2+3n_3+4n_4+7n_7$, the baryon number density at fixed volume. In particular the mass-seven capture produces two helium-4 nuclei, not one.

The beryllium-to-lithium step is electron capture. Eliminating beryllium assumes that its capture flux tracks its production on the slow evolutionary timescale. More generally one retains ${\dot{n}}_{\mathrm{Be}}=R_{34}-{\lambda }_e{n}_{\mathrm{Be}}$ and ${\dot{n}}_7={\lambda }_e{n}_{\mathrm{Be}}-R_{17}$; setting the former to zero gives the effective equation used above. Fast capture relative to slow evolution alone would not prove that lithium exceeds beryllium: if both reach steady state their ratio is $n_{\mathrm{Be}}/n_7={\lambda }_{17}{n}_1/{\lambda }_e$. The mass-seven abundance assertion is thus part of the stipulated schematic limit, not a consequence to impose on every detailed solar model.

The enormous separation between the one-second deuterium destruction time and the other stated timescales justifies deuterium quasi-equilibrium in proton-proton burning:Substitution givesThe approximation is to the rapidly adjusting intermediate abundance, not to the slow proton abundance.

Near the centre the helium-3 relaxation time is also short compared with solar age and with the timescale of significant hydrogen evolution. Put $A={\lambda }_{33}$, $B={\lambda }_{34}{n}_4$ and $C={\lambda }_{11}{n}_1^2/2$. For slowly varying background quantities, the stable positive root of $A{n}_3^2+B{n}_3-C=0$ gives the helium-3 equilibrium abundanceThe other root is negative and unphysical. The production-minus-destruction function decreases strictly with positive $n_3$, so this is the unique attracting equilibrium. For $n_3=n_{3e}+x$, with the background held fixed during relaxation,Dropping the quadratic perturbation term gives the helium-3 relaxation timeIf the equilibrium itself evolves slowly, an additional forcing term $-{\dot{n}}_{3e}$ appears; tracking is accurate when this drift is small over one relaxation time. The given central value $6\times 10^5$ years is much shorter than the roughly $4.6\times 10^9$-year age of the Sun.